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  • What is the value of <img alt="\lim _{x \rightarrow \frac{\pi}{2}}\left(1+x-\frac{\pi}{2}\right)^{1 / \sin \left(\frac{\pi}{2}-x\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%

What is the value of \lim _{x \rightarrow \frac{\pi}{2}}\left(1+x-\frac{\pi}{2}\right)^{1 / \sin \left(\frac{\pi}{2}-x\right)}?

Option: 1

1


Option: 2

0


Option: 3

\frac{1}{e}


Option: 4

e


Answers (1)

best_answer

Given: \lim _{x \rightarrow \frac{\pi}{2}}\left(1+x-\frac{\pi}{2}\right)^{1 / \sin \left(\frac{\pi}{2}-x\right)}
If \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$, then $e^{\lim _{x \rightarrow a} \frac{f(x)}{g(x)}}.
=e^{\lim _{x \rightarrow \frac{\pi}{2}} \frac{x-\frac{\pi}{2}}{\sin \left(\frac{\pi}{2}-x\right)}}

Use the trigonometric formula \sin \left(\frac{\pi}{2}-x\right)=\cos x.

=e^{\lim _{x \rightarrow \frac{\pi}{2}}\frac{\frac{2x-\pi}{2}}{\cos x}}
=e^{\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 x-\pi}{2 \cos x}}

Applying L-Hospital's rule, the above expression becomes,

=e^{\lim _{x \rightarrow \frac{\pi}{2}-2 \sin x} \frac{2}{}}
=e^{\frac{2}{-2}}
=e^{-1}
=\frac{1}{e}

Therefore, \lim _{x \rightarrow \frac{\pi}{2}}\left(1+x-\frac{\pi}{2}\right)^{1 / \sin \left(\frac{\pi}{2}-x\right)}= \frac{1}{e}.

Posted by

vishal kumar

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